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Question about braking force

  1. Sep 30, 2011 #1

    Mvb

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    Hello
    I am having some trouble solving a problem related to braking forces. The assignment is this:

    A birdwatcher sees an egg falling from a 20 m tall tree. (Though the 20 m might not have anything to do with this particular task because it is part of a longer question).
    The egg strikes the ground and is brought to a stop in a distance of 1 mm. Assuming a mass of 20 g for the egg calculate the force required. (You may assume a constant braking force. Use g = 10 m/s^2)

    I was going to calculate the eggs force (F = 0,02 kg * 10 m/s^2 = 0,2 N)
    Because in order for the egg to become stationary, the forces on it must be equal in magnitude and have opposite directions (and this is the force on the egg towards the earth).

    I am not sure what to do about the 1 mm though?

    Thanks for your help
     
  2. jcsd
  3. Sep 30, 2011 #2

    gneill

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    Staff: Mentor

    Hi Mvb, welcome to Physics Forums.

    Please try to follow the posting template when you start a homework thread :smile:

    The 10 m/s^2 acceleration alone applies as the egg is falling, causing the egg to speed up as it falls. It will have some final velocity just before it contacts the ground.

    After it makes contact with the ground you're told that it comes to a halt in a distance of 1mm. So that means it goes from that final velocity to zero in that short distance. That represents a different acceleration rate which applies over that distance.
     
  4. Oct 1, 2011 #3

    Mvb

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    Thank you, and I'll do that the next time.

    I understand that I am supposed to find the acceleration that makes the egg go from 20 m/s to 0 m/s in 1 mm, but I'm not sure how to do that?

    I graphed it and was thinking to find the acceleration like this:

    a = (0 m/s - 20 m/s) / Δt
    0.001 m = 1/2 * a * t^2 + 20 m/s * t

    And solve.
    But I'm not supposed to use a calculator and it seems unlikely that it is that complicated.
     
  5. Oct 1, 2011 #4

    gneill

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    Staff: Mentor

    There's another kinematic formula involving acceleration and velocities (initial and final) that involves distance rather than time. It would be the easier route to go. For a body undergoing acceleration from an initial velocity vi to a final velocity vf over distance d,
    [tex] v_f^2 = v_i^2 + 2 a d [/tex]
     
  6. Oct 1, 2011 #5

    Mvb

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    Thank you very much, that was really helpful :)
     
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